18
$\begingroup$

How long a series of points in (0,1) can be chosen such that the first two are in different halves, the first three are in different thirds, ... the first $n$ are in different $n^{\text{th}}$s? My first try of $(0+,1-,\frac{1}{2}-,\frac{3}{4}-, \frac{1}{5}+,\frac{5}{8}-,\frac{1}{3}-,\frac{7}{8}-,\frac{1}{3}+)$ works through $9$, but there are two points in $(0.3,0.4)$. The plus and minus signs indicate a shift of some distance away from the given point small enough not to move over any fraction of interest.

$\endgroup$
1
14
$\begingroup$

An answer is asserted but not proved at the following URL:

http://en.wikipedia.org/wiki/Irregularity_of_distributions

$\endgroup$
3
  • 1
    $\begingroup$ That is very surprising. I'm glad I didn't spend more than a few minutes on this! I never would have reached anything close to 17. $\endgroup$
    – Sputnik
    Jun 5 '11 at 1:59
  • $\begingroup$ Fascinating. Interestingly, the solution roughly matches Ross' up to $5/8-$. $\endgroup$
    – joriki
    Jun 5 '11 at 1:59
  • 8
    $\begingroup$ See also mathoverflow.net/questions/19896/… and the freely available original proof at citeseerx.ist.psu.edu/viewdoc/… $\endgroup$
    – joriki
    Jun 5 '11 at 2:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.